Question

Suppose f(x)=x6+3x+1f(x)=x6+3x+1. In this problem, we will show that ff has exactly one root (or zero) in the interval [−4,−1][−4,−1].

(a) First, we show that f has a root in the interval (−4,−1)(−4,−1). Since f is a SELECT ONE!!!! (continuous) (differentiable) (polynomial)   function on the interval [−4,−1] and f(−4)= ____?!!!!!!!

the graph of y=f(x)y must cross the xx-axis at some point in the interval (−4,−1) by the SELECT ONE!!!!!! (intermediate value theorem) (mean value theorem) (squeeze theorem) (Rolle's theorem) .Thus, ff has at least one root in the interval [−4,−1]

(b) Second, we show that ff cannot have more than one root in the interval [−4,−1] by a thought experiment. Suppose that there were two roots x=a and x=b in the interval [−4,−1] with a<b. Then f(a)=f(b)=____? Since f is SELECT ONE!!!!!! (continuous) (differentiable) (polynomial)

on the interval [−4,−1] and SELECT ONE!!!!! (continuous) (differentiable) (polynomial) on the interval (−4,−1)  by SELECT ONE!!!!!! (intermediate value theorem) (mean value theorem) (squeeze theorem) (Rolle's theorem) there would exist a point c in interval (a,b) so that f′(c)=0. However, the only solution to f′(x)=0 is x=____?!!!!!!which is not in the interval (a,b), since (a,b)⊆[−4,−1] Thus, f cannot have more than one root in [−4,−1].

Fill in the blanks with question marks

select one of the answers where it says (SELECT one!!!!!!!!)

Thank you

Answer 1